MCPCBs for LED Applications - - Get a Free Blog Here
Transcript of MCPCBs for LED Applications - - Get a Free Blog Here
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MCPCBs for LED ApplicationsThermal Management Material Specifications
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To break down your needs for thermal management materials, specifically metal core printed circuit board (MCPCB), in LED Applications by technical requirements in order to make more effective callouts.
Controlling the processThe end result should be shorter lead times, lower cost, and more reliable product.
Non-browning soldermasksIntroduce new non-browning white LED soldermasks.
Purpose of the WebinarDefining your needs
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The MCPCB commonly consists of a metal core layer (typically aluminum or copper), a continuous dielectric layer and a copper circuit layer.
Metal Core BoardsDefinition
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Part OneSpecifying Materials for LED Applications
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Survey QuestionHow do you specify materials in MCPCB LED applications?a) Company Name and/or Part Number (e.g. Bergquist, Laird, DuPont, etc.)b) Thickness of dielectricc) Tg or Td
d) “Metal Core”e) Insulation Resistance & Thermal Conductivity Values
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Survey Answer
CommentSpecifying by brand name locks you into a particular product produced by a particular manufacturer.
Risks Include:• Locked into pricing• Subject to lead times• Preventing new materials from being used on your product
How do you specify materials in metal core LED applications?
a) Company Name and/or Part Number (e.g. Bergquist, Laird, DuPont, etc.)
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How do you specify materials in metal core LED applications?
b) Thickness of dielectric, c) Tg, d) Td
Survey Answer
Specifying only these items may not fully address your needs.• Does not address thermal conductivity• Does not address electrical insulation resistance• Does not address type of dielectric
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Comment: Specifying “metal core” does address anything whatsoever.
Answer
How do you specify materials in metal core LED applications?
e) “Metal Core”
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Comment: You are a rock star! This is the whole point of our webinar.
Answer
How do you specify materials in metal core LED applications?
f) Insulation Resistance & Thermal Conductivity Values
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Consider your needs when bringing the bare board into the equation
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Survey QuestionWhat Are Your Needs When Selecting Materials? (Choose all that apply)a) Transfer heat (Thermal Conductivity)b) Prevent short circuiting to base metal core (Electrical Insulation Resistance)c) Thickness of dielectricd) Brand name of the material?
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The # 1 cost component of the MCPCB is the dielectric substrate between the copper traces and the metal heatsink / core.
Cost Drivers For MCPCBsDielectric Substrate
An insulating medium which occupies the region between two conductors. In this case, the copper circuits and the metal core heat sink.
Definition
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The most effective way to reduce cost of dielectric is to introduce competition: Laird Bergquist Dupont Uniplus Iteq Insulectro Plus future innovators
Cost Drivers For MCPCBs (cont.)
Introducing competition
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Definition and Applications
Thermal Transfer
Product Application
Electrical Insulation
Product Reliability
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Common Callouts include Thermal Impedance / Resistance (°C in2/W) and Thermal Conductivity / (°W/m-K).
Purpose #1Transfer Heat
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Heat Resistance
Stability
Flexibility
Dielectric Properties
Density/Weight
CooLamTM MCPCB (metal core PCB) Offers:• Very Low Thermal Impedance• Excellent Reliability Performance • Excellent Durability and Stability at High Temperature• Uniform Thermal, Mechanical & Electrical Properties Under Environmental Stress • Lead Free Solder and Wirebond Process Compatibility• Halogen Free• Meets UL 94 V-0• Construction Variations to Meet Thermal Management Needs• 3D Shapes
Why Kapton® / Polyimide?
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Objective:Measuring thermal performance per ASTM D5470.
Equipment:1) Steady State “Thermal Interface Material Tester” (TIM)
(Analysis Tech. Inc.)
Factors to Consider:1) Reducing contact resistance between sample and test unit2) Repeatability of measured values3) Identify any equipment and/or material limitations
Thermal MeasurementCharacterize the Thermal Performance of Materials
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Basic Principles of ASTM D5470T1
T2
T3
T4
Power (Watts)
Heat Source
Heat Sink
What is Thermal Resistance ?
Thermal Resistance (Rth) is defined as the difference in temperature between two closed isothermal surfaces divided by the total heat flow between them.
Rth = (T(A) – T(B)) Power
(surface B)
(surface A)
∆C
Watt (V*I)
* Present system does a good job of accounting for all heat and monitoring temperature but nothing is perfect.
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Basic Principles of ASTM D5470 (cont.)
T1
T2
T3
T4
Power (Watts)
Thermal Resistance (Rth): ∆CWatt
= (∆ Temp. power)
Thermal Impedance (RthI): C - in2
Watt= (Thermal Resistance) x area (of test sample)
Thermal Conductivity: Watt
m -C
= thickness (material) thermal impedance
Test Sample
and
Output from TIM unit:
Heat Source
Heat Sink
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ASTM D5470 Thermal Measurement Technique
Test Sample
“thermal grease”
“thermal grease”
Power (Watts)
Copper
Dielectric
Aluminum
• Total Thermal Resistance (Rth total) is the sum of components and its thermal resistance value.
A B C
Grease
Grease
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ASTM D5470 Thermal Measurement Technique
TestSample
“thermalGrease”
“thermalGrease”
RthI (grease_top)
RthI (sample)
RthI (grease_bottom)
+
+
Total RthI
Measured Thermal Impedance (C-in2/Watt):
(RthI (grease_top)RthI (sample) = RthI (grease-bottom))+-Total RthI
Total RthI
AB
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k
thicknessimpedancethermal
Electronics industry defines thermal impedance as the following:
This does not include an area through which the heat flows!
Thermal Impedance
Diel
Diel
k
tI
W
KmmI
mKW
2
Units:
Ak
tR
Diel
Diel
W
K
m
mR
mKW
2
Units:
Thermal Resistance
Thermal Impedance
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Thermal Impedance
0.0050.0040.0030.0020.001
0.225
0.200
0.175
0.150
0.125
0.100
0.075
0.050
Thickness (in)LC
Th
erm
al Im
pe
da
nce
K
-in
^2
/W
S 0.0010461
R-Sq 100.0%
R-Sq(adj) 100.0%
Fitted Line Plot K-in^2/W = 0.01366 + 39.35 Thickness (in)
Data Plotted from multiple readings
Calculating the thermal conductivity of a materialPlot Ra vs. thickness from TIM tester1/k =thermal conductivity k = is the slope of the line
Thermal Conductivity LG: 0.25 W/M-K
= 1.0 in / 39.35 K-in2/W = .006394 W/in-K x 39.4 in/M (convert to metric)= 1.0 W/M-K
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Example problemsWhat is the thermal resistance through the thickness of a 0.3 meter by 0.3 meter piece of stainless steel that is 1.5 cm thick? Assume the thermal conductivity of stainless steel is 15 W/mK.
0.3m
0.3m
1.5cm
kA
tR
)3.0)(3.0)(15(
015.0
mmmKW
m
kA
tR
WKR 011.0
k=15 W/mK
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Example ProblemsThis same piece of stainless steel now has a heater attached to one side that generates 100 Watts of heat into the plate.
If the bottom of the plate measures 45°C (318K), what is the temperature of the top side of the plate?
thermal
bottomtop
R
TTq
)( thermalbottomtop RqTT )(
bottomthermaltop TRqT
KWKWTtop 318011.0100
CKTtop 1.461.319
0.3m
0.3m
1.5cm
k=15 W/mK
Ttop
Tbottom
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Basic Question
Al/Cu Base
LED
Heat Sink
CuDielectric
Tjunction = 65°C
Al/Cu Base
LED
Heat Sink
CuDielectric
Same LED and power input
Room Temp = 30°C
MCPCB 1
MCPCB 2
Tjunction = 70°C
WHY???
Heat Transfer
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Thermal Management for LED Applications
Presented by Clemens Lasance
Clemens is a former Principal Scientist Emeritus with Philips Research, the Netherlands, with a 30 year +
focus on thermal management of electronic systems.
He is now a consultant for Somelikeit Cool, contact info: [email protected]
What is the role of the PCB?
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MotivationProviding the right information the first timeThe goal is to provide the LED application engineer with the right information to select the mostoptimal MCPCB from a thermal point of view. Thus, ensuring the best decision is made for aspecific application in regards to thermal performance and cost.
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Reasons / Key Issues
Lifetime
Color point
Efficiency
Main Goal of Thermal ManagementThe Calculation of application-specific junction temperatures
All these key issues are significantly dependent on the junction temperature.
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Determining critical temperatures is contingent upon a critical understanding of:
• The electrothermal analogue• Thermal conductivity k• Heat transfer coefficient h• Thermal resistance Rth
Basics of Heat TransferFundamentals of Critical Temperature Calculation
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∆T = q * Rth ∆V = I * R
Temperature drop (°C) Voltage drop (V)
Power dissipation (W) Current (A)
Heat flux (W/m2) Current density (A/m2)
Thermal resistance (K/W) Resistance (Ohm)
Electro-Thermal Analogue
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Electro-Thermal AnalogueA heat transfer path can be described by an electrical network
T ambient
T source
q
Rth
ΔT = Tsource-Tambient
Rth (K/W) = thermal resistance
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T ambient
T source
q
R1R2
T intermediate
Series CircuitExample for two resistances in series
DT = Tsource - Tambient = q * Rtotal
Rtotal = R1 + R2
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T ambient
T source
q
R1 R2
T intermediate
Consequence of Series CircuitOften only the largest R is relevant!
Designer’s job: find the largest resistance!
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hot
Heat insulation
cold
L
Cross Sectional Area A
Heat flow q
Heated block Water cooled block
TL
Akq D
Thermal Conductivity: Fourier’s Experiment (1822)
Result of Fourier’s experiment:q ~ ∆T = Thot – Tcold
q ~ A = cross-sectional areaq ~ L-1
The proportionality constant k is called the ‘thermal conductivity’
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Typical Thermal Conductivity Values
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TAhq D
The Heat Transfer Coefficient hThe heat transfer from a solid wall into a fluid (air, water) is called convection, and to first order this heat transfer is proportional to the area and the temperature difference between the wall and the fluid:
The proportionality constant h is called the ‘heat transfer coefficient’
Typical values:Natural convection: 10 W/m2KForced convection: 50 W/m2K
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Information required for first order guess:
• Maximum junction temperature: e.g. 120 °C
• Maximum ambient temperature: e.g. 40 °C
• An estimation of all thermal resistances in the total heat transfer path
Suppose a Designer needs 5W to reach a required light output
What procedure is most optimal?
Practice
totalthR
Tq
D
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Practice (2)
Next steps
Check if required power is manageable by suitable heat sinks & convection
Check which thermal resistances are dominant
Then focus on them to reduce the total thermal resistance
Final step should always be a detailed analysis
Recall that often we don’t talk one-dimensional heat transfer but heat spreading, which is a rather complicated issue
For more information please see Heat Spreading, Not a Trivial Problem in the May 2008 issue of ElectronicsCooling
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Practical Example
Relevant thermal resistances, assume 1 cm2 PCB
K/W
Rth-LED 16 Luxeon Rebel
Rth-MCPCB 0.5 t = 100 µm, k = 2 W/mK
Rth-TIM 1 t = 100 µm, k = 1 W/mK
Rth-heatsink 50 A = 20 cm2, h = 10 W/m2K
Rth-heatsink 2 A = 100 cm2, h = 50 W/m2K
Rth-heatsink 1 A = 1 cm2, h = 10000 W/m2K
TL
Akq D
Ak
LRth
TAhq DAh
Rth
1
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Practical ExampleWe need 5W for our light outputCan we find a suitable heat sink?
We have:
Tjunction = 120 °C
Tambient = 40 °C
Rth-LED + Rth-MCPCB + Rth-TIM = 17.5 K/W; Rth-heatsink = ?
Can we dissipate 5W? No way:
q = 80 / (17.5 + Rth-heatsink)
Conclusion: Even with an ideal heat sink (Rth-heatsink = 0) we cannot dissipate the required 5W. In this case we need an LED with less thermal resistance.
totalthR
Tq
D
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totalthR
Tq
D
Practical ExampleWe need 1W for our light output
We have again:
Tjunction = 120 °C
Tambient = 40 °C
Rth-LED + Rth-MCPCB + Rth-TIM = 17.5 K/W; Rth-heatsink = ?
Which heat sink is OK?q = 80 / (17.5 + Rth-heatsink) ; Rth-heatsink = 62.5 K/W
We found earlier:A=20 cm2, h=10 W/m2K Rth-heatsink = 50 K/W
Conclusion: We can use a simple heat sink with natural convection.
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Thermal resistance of the MCPCB does not play any role
In our first case, the LED thermal resistance was dominant; while in the last case, the convective resistance was dominant.
Hence, from a thermal point only, you can choose MCPCBs with a much lower thermal performance and, hence, lower cost.
The thermal performance of the PCB is relevant only for very high heat flux cases (e.g. liquid cooling) and for top-of-the-bill LEDs.
Practical Example Conclusion
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NomenclatureThe confusing situation regarding ‘thermal impedance’
Is this a problem? Yes, because time-dependent (dynamic) test methods will be increasingly used, one output of which is the ‘correct’ thermal impedance.
Proposal:Use thermal resistance per unit area, or unit Rth.
Fact: ‘Electrical impedance’ is historically reserved to describe time-dependent electrical resistance. In the limit of steady state, thermal impedance equals thermal resistance hence, units should be the same!
Hence,
‘Thermal impedance’ , as used by U.S. vendors, violates the electro-thermal analogy, because:
– Unit does not correspond (K/W vs. m2K/W)
– Definition does not correspond (time-dependent vs. steady state)
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Electrical Properties
Presented by Michael Gay (Isola) and Norm Berry (Insulectro)
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• Insulation Resistance (IR)• This test is used to provide a quantifiable resistance value for all of a product's insulation
• Dielectric Breakdown Strength Test• The test voltage is increased until the dielectric fails, or breaks down, allowing too much
current to flow.
• Dielectric Withstand Voltage Test (DWS) or High Potential Test (HiPot) • A standard test voltage is applied (below the established Breakdown Voltage) and the
resulting leakage current is monitored
• Conductive Anodic Filament Test (CAF)• This test is used to determine the capability of the material for various design feature spacing
under high humidity conditions with applied voltage
Common Tests
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Proof Test or High Potential Test (HiPot) A standard non-destructive test voltage is applied (below the established Breakdown voltage) and the resulting leakage current is monitored. The test is done to evaluate the integrity of the dielectric material.
Typical Proof Test (VDC)
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High Potential Dielectric Testing (HiPot) This non-destructive testing methodology is used to characterize the laminate material insulation capability under a specified potential. Leakage current is measured and compared to predetermined limits
Typical Test Conditions:500-2500 Volts DC (VDC)700-1200 Volts AC @50-60 Hz (VAC)Up to 30 seconds10 MΩ – 500 MΩ10 μA – 100 μA
-
+Applied
Potential
Typical Proof Test (VDC)
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Breakdown voltage is the stress at failure when AC at power frequency is applied using a rate of rise is 500 Volts/second. Breakdown voltage testing does not relate to proof stress “Proof Testing”. Breakdown voltage is the potential difference at which dielectric failure occurs under prescribed conditions in an electrical insulating material located between two electrodes.
Breakdown (kV AC)Breakdown Voltage Test
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This testing methodology is used to characterize the laminate material insulation capability to failure.
Typical Test Conditions:Ramp Volts AC @50-60 Hz (VAC) until failure
Example: Determine the electric strength in volts per mil for each specimen by dividing the breakdown voltage expressed in kilovolts by the thickness express in inches. -
+Applied
Potential
ES = (6.8 KV/.005 inch) X (1000 V/KV) X (1 inch/1000 mils) = 1360 v/mil
Breakdown (kV AC)Dielectric Breakdown Voltage
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Typical thermal management materials are modified to have higher thermal conductivity. The modifications of the dielectric material raise the Dk of the material to the range of 6 or higher.
Permittivity (Dk)
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Product ReliabilityPresented by Richard A. Wessel of DuPont
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Al/Cu Base
LED
Heat Sink
CuDielectric
Tjunction
Tsolder point
Tboard
Tambient
q
Tjunction
Tsolder point
Tboard
Tambient
RJ-SP
RSP-B
RB-A
Big concern: what is Tjunction?
Simplified LED
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Lower T junction Better
Cree® XLamp® MC-E LED
Luxeon® Rebel
Increased relative luminous flux for lower junction or thermal pad temperature
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Increased LifetimesLower Junction Temperature
Luxeon® Rebel
Increased lifetimes for lower junction temperature
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Suggested Fabrication NotesTaking Control of Your Design!
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Take your requirements for each characteristic that you deem most important and specify (values for thermal conductivity and electrical insulation resistance).
We also suggest including a list of pre-approved materials.
Suggested Fabrication NotesPurpose of Fab Notes
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Copper Weight: 2 oz.
Metal Core: 0.060” Al 5051
Dielectric:
Thermal Conductivity: Must meet or exceed 1.3 (°W/m-K)
Thermal Resistance: Maximum 0.9 (°C in2/W)
Electrical Insulation: Must meet or exceed 2,000 VDC
Any other characteristic you feel is important: XYZ
Currently Approved Materials: Bergquist HT-07006 Bergquist LTI-06005 Laird T-lam SS 1KA06 DuPont CooLam
Suggested Fabrication Notes (cont.)
Example of a Fab Note
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Allows supplier to quote multiple vendors.
Subsequent Results• Lower Cost• Quicker Lead Times• Adaptability to take advantage of new technologies down the road• Introduce competition from the raw materials and finished PCB suppliers
Suggested Fabrication Notes (cont.)
Benefits of Fab Notes
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Materials Chart
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Part TwoNon-browning White Soldermask
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Focus on initial thermal shock, not constant heat
Tend to Brown over time
Not developed to be a Reflecting Surface for LEDs
White SoldermaskCurrent Resin Systems
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Remain white through Assembly and Lifespan of LED final product
Have high reflectivity to enhance LED performance Some believe high reflection soldermask results in lower power consumption for same light output
White Soldermask (cont.)
Desired Resin Systems
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Past two years a number of “LED” white soldermasks have been introduced Unfortunately we have not seen improvements in all of them
Through numerous studies we have so far found two offerings that show significant improvement: Peters (SD2491SM TSW) Sun Chemicals (CAWN 2589/2591)
What are the Solutions?New White High Temperature Liquid Photo-Imageable Soldermasks
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If Bright white soldermask is the objective, these solutions may not suffice
Suggest base coat of LPI White then topcoat of High-Temp thermal white soldermask
After numerous thermal whites, one does not brown after multiple cycles (Peters SD2496TSW)
This is the formula behind “proprietary” white soldermasks offered by other PCB vendors
What are the Solutions? (cont.)
New White High Temperature Thermal Soldermasks
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To date, we’ve seen cost increases over standard soldermask price of 30% - 100% Approximately .20 to .30 cents per panel (18” x 24”)
New White High Temperature Thermal SoldermasksTo date, we’ve seen cost increases over standard soldermask price of > 100%
Approximately $1 per panel (18” x 24”) Includes labor and set up for secondary application process
Cost ExpandersNew White High Temperature Liquid Photo-Imageable Soldermasks
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your needs for thermal performance of the dielectric
your needs for the electrical performance of the dielectric
your needs in context of entire product, not just PCB (when applicable)
that you have white soldermask choices
but keep your fab notes open to take advantage of future developments
Thermal Webinar Conclusion
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